Can a Capacitor be fully charged?

Status
Not open for further replies.

Corky

Member
Hi all,

i know practically a cap can be fully charged (to an accuracy we take as 100%) but as the voltage goes up in a cap the current drops equally so even when the cap is 99.9999999999% charged would the current would keep lowering so the cap would never actually reach full charge? or would it eventually reach its max?

Regards
 
I would guess noise on the power lines would mean it goes above and below the average supply voltage.

Mike.
 
THEORETICALLY, no, the capacitor will not reach the supply voltage when charged through a resistor.

but

PRACTICALLY, after a period of about five time constants you are as near to "fully charged" as makes no difference.

JimB
 
HI Jim,

yer thats what i meant the Theoretical part, just curious!!

even with infinite time?

Regards
 
This is quite easy to show theoretically

https://www.wolframalpha.com/input/?i=exp(x) <- shows that the e^x function will never reach 0 even with infinite time.
So as Vc = Vin (1-e^(-t/RC)), Vc will never be equal to Vin.

I should add that even without a series resistor you'll still have the ESR of the capacitor, so the only capacitor that would be able to actually ever reach the supply voltage (theoretically) would be an ideal capacitor. A super capacitor would come closest.

However as everyone else has said in practice PSUs have ripple and due to the accuracy of the measurement its not noticeable.
 
Last edited:
Given infinite time - yes it get there! No problem.
You just have to wait a while.

JimB
 
Mmmm...
A disagreement from Misterbenn.
I can agree with that.

I was making the (possibly bad) assumption that (1/infinity) = 0

JimB
 
I think it can. Because even if e^x can never reach 0 becoming smaller and smaller, charge of real capacitor consists of electrons and you can't have charge less than one electron's charge. So when the "last" electron will go to cap, if next electron will try to move to cap, voltage on cap will be bigger than supply voltage, so it won't do that, cause electric field of other electrons in cap won't allow him, so we can say, that cap is fully charged. But in real life we haven't power supplies with one electron charge voltage precision, so there is always small current trough capacitor caused by power supply fluctuations.
 
Hi everyone,

the way Nyak was thinking is where i am now, the 'last' electron(to make the supply voltage) will eventually move and will eventually have to get to the cap how ever slow. so in a perfect world the cap would be fully charged at a some point the only time electrons would stop is when there would be no attraction between the plates i.e. supply voltage?

is this correct?

Regards
 
In an ideal world where there was absolutely no resistance and no leakage yes you may be right.

But you have to consider at least the ESR of the capacitor. So that last electron will see some impedance but the attraction (potential difference) will be miniscule, so does it bother to make the trip over to the capacitor plates, probably not. WE might even have to start thinking of quantum theory when getting to this kind of level and if you want to cover all the bases.

But I think its good to remember the second law of thermodynamics, which predicts that given infinite time all systems will have equilibrium and entropy will be zero. ... so in that respect you could say the capacitor voltage and supply voltage will eventually be equal.
But at this point we're starting to sound more like the answer to an Oxford entrance exam...
 
i know what your'e saying but unfortunately thats how my brain works, it really cant just accept something and thats why i end up asking questions like this, i know it will never matter to me wether it Charges over 99% but its always nice to know, at least then i can sound like i know what im talking about on my oxford entrance exam haha.

Cheers guys, just here to get the brains working

P.s. my next message is my 100th so i have to make it a good one (its a big milestone for some one who was only going to use this site for one question/answer)
 
Is it possible to argue over 99.9999999999%, for ever?
No but almost.
lol
----edited-----
Corky,
Congratulations on 99 almost 100.
Maybe you can write a very short post (0.9).
99.9
 
Sometimes it is difficult to decide when some effect is so small that it does not make any measurable difference.

Consider a vehicle, say a 40 tonne truck going down the road at 100kph.
It hits a big flying insect with a mass of 1 gram flying towards the truck at 1kph.
The principle of conservation of momentum tells us that the truck will slow down due to the collision.
But measuring the change in velocity is a rather difficult, going on impossible, task.

And so it is with that "last electron" lost in the twilight zone!

JimB
 
But we cant deny it has happened,

for my 100th message im going to say rip to all the bugs who people have marked down to not making a real difference to car speed

speed kills!!
 
speed kills!!
Click to expand...
Certainly does for the bugs!

And just for fun I calculated the change in speed of the truck mentioned above, it slows down by 2.523999942 millimetres per hour. (always assuming that I have not dropped a power of 10 or two).

JimB
 
haha, well thats my work day finished and i believe i have achieved something, 100 messages not bad

well done JimB wealth of knowledge
 
A related joke about the 99.99999999%

Two individuals, a mathematician and an engineer are on the far end of a room.
On the other end, there is a pot of gold.
A lamp genie comes in and says:
"Either one of you can take the pot of gold, but there is one condition: Your first step will be across half of the room. The next step, half of that half, the third step yet another half of the previous halves. And so on".
The mathematician starts crying. He knows that to reach the pot of gold, it will take him an infinite amount of time.
But the engineer starts smiling: "Theoretically, I will never reach the pot of gold. In practice however, I can get close enough to grab it".

And that my friends, is the difference between theory and practice. In other words, between pure science and engineering.
 

If you mean charged with electrons, a capacitor contains the same net number of electrons at zero volts as it does at 100 volts. This is because for every electron that accumulates on one plate, another electron depletes from the opposite plate for a net gain/loss of zero. The plates have an unbalance of electrons, but the net electron gain/loss is zero. Therefore it is wrong to say a capacitor is "charged" when no change has occurred. You should say "energized" instead, because a cap with more voltage across its terminals contains more energy.

Ratch
 
I like this answer best. But, as Ratch says, it's the transfer of charge from one plate to the other. I know you know that, but the wording was misleading. However, unlike Ratch, I have no problem with using the word "charged" to describe what happens to a capacitor. We should all know what the word means. Charging, after all, always involves moving charge from one place to another, in some sense. Try saying "energized" ... and no matter how "correct" that might be, you are more likely to confuse people. There are many ways to energize something, including heating it, throwing it, or attaching it to a sling shot prior to launching it at a house to cause some mischief.
 
Status
Not open for further replies.

Similar threads

  • Question
What can be a cheaper alternative to IC 5L0380R?
Replies
14
Views
3K
electronium
Discharging A Charged 10,000 Volt Capacitor
Replies
7
Views
2K
Tony Stewart
  • Question
Can the amperage of this power supply module be modified to increase?
Replies
12
Views
2K
schmitt trigger
S
V
  • Question
How to get SOF, EOF, CRC etc of a CAN Frame.
Replies
8
Views
1K
Papabravo
  • Question
How can a clamp meter be used to measure the amperage of a wire or cable from the phase and neutral of two strands?
Replies
6
Views
2K
Nigel Goodwin
Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…